Given figure is a graph of the electric potential in a region of gravity free space where electric field is parallel to the x-axis. A charge q of mass m is projected from origin along positive x-axis

Question

Given figure is a graph of the electric potential in a region of gravity free space where electric field is parallel to the x-axis. A charge q of mass m is projected from origin along positive x-axis with speed v 0 ​ then ( m q ​ = 2 ) :-

TestHub · Accepted Answer

By w.e.t 2 1 ​ m v 2 − 0 = q [ v F ​ − v 1 ​ ] If particle will reach at x = 6 . It will difinately cross x = 8 ​ v 2 = m 2 q ​ [ 14 − 5 ] ​ v = 2 × 2 × 9 ⇒ v = 6 m / s ​ For x = 4 It should cross x = 1 ​ v ≥ 4 m / s ​ v f 2 ​ = v l 2 ​ + 2 m q ​ ( v f ​ − v I ​ ) ​ Maximum velocity is at x = 3 ∵ v f ​ ​ = 4 v ​ = 64 + 2 × 2 ( 9 ) v f ​ ​ = 10 m / s ​

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